A factory manufactures two types of screws, A and B, each type requiring the use of two machines - an automatic and a hand - operated. It takes 4 minute on the automatic and 6 minutes on the hand - operated machines to manufacture a package of screws ‘A’, while it takes 6 minutes on the automatic and 3 minutes on the hand - operated machine to manufacture a package of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws ‘A’ at a profit of 70 P and screws ‘B’ at a profit of ₹ 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
Let the factory manufacture x screws of type A and y screws of type B on each day,
Therefore, x 
0 and y 
0.
The given information can be compiled in a table as follows
4x + 6y 
240
6x + 3y 
240
The manufacturer can sell a package of screws ‘A’ at a profit of Rs 0.7 and screws ‘B’ at a profit of Re 1.
Total profit, Z = 0.7x + 1y
The mathematical formulation of the given problem is
Maximize Z = 0.7x + 1y
subject to the constraints,
4x + 6y 
240
6x + 3y 
240
x, y 
0
First we will convert the inequations into equations as follows:
4x + 6y = 240, 6x + 3y = 240, x = 0, y = 0.
Region represented by 4x + 6y 
240
The line 4x + 6y = 240 meets the coordinate axes at A(60,0) and B(0,40) respectively. By joining these points we obtain the line 4x + 6y = 240. Clearly (0, 0) satisfies the 4x + 6y 
240. So, the region which contains the origin represents the solution set of the inequation 4x + 6y 
240.
Region represented by 6x + 3y 
240
The line 6x + 3y = 240 meets the coordinate axes at C(40,0) and d(0,80) respectively. By joining these points we obtain the line 6x + 3y = 240. Clearly (0, 0) satisfies the 6x + 3y 
240. So, the region which contains the origin represents the solution set of the inequation 6x + 3y 
240.
Region represented by x 
0, y 
0 :
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x 
0 and y 
0.
The feasible region determined by the system of constraints 4x + 6y 
240, 6x + 3y 
240, x 
0,
y 
0 are as follows.
The corner points are C(40,0), E(30,20), B(0,40). The values of Z at these corner points are as follows
The maximum value of Z is 410 at (30, 20).
Thus, the factory should produce 30 packages of screws A and 20 packages of screws b to get the maximum profit of Rs 410.